[Axiom-developer] Re: Integral challenge

[Waldek Hebisch]

Bob McElrath wrote:
> So, any integral experts want to take a crack at it?  In the problem I
> am interested in, a is real and a>0, b is real and 0 <= b <= 1, and I
> need the integral over theta=0..2*Pi, in case the definite integral is
> easier.
> 
> (3) -> integrate(exp(-a*sqrt(1-b*cos(theta))), theta)
>  3) ->
>    >> Error detected within library code:
>   Function not supported by Risch d.e.
<snip>
> But, I'm surprised to find that such a simple integral has no known
> solution.

>From Axiom point of view your function is very complicated: you have
a transcendental function, that is "cos(theta)" and then you make
an algebraic extension ("sqrt(1-b*cos(theta))") and finally you
again make a transcendental extension. 

There is a calculs-102 method to simplify your integral: take x = cos(theta).
Then we get a new integral:

integrate(exp(-a*sqrt(1-b*x))/sqrt(1 - x*x), x)

Now we have a transcendental extension of an algebraic extension. On this
one Axiom just runs out of memory. However this one is still pretty
complicated since we have an algebraic function in the exponent. So
put z = sqrt(1-b*x). Now we get:

integrate(z*exp(-a*z)/sqrt(1 - ((1 - z^2)/b)^2), z)

which Axiom returns unevaluated. AFAIU that means that Axiom proved
that the function have no elementary integral. 

One way to handle the integral above would be to expand exponential
into Taylor series. Then we will get a sequence of integrals like

integrate(z^k/sqrt(1 - ((1 - z^2)/b)^2), z)

Those are elliptic integrals, typically non-elementary. Since thay have
a common irrational part one can reduce then to a couple of base
cases. Of course, the tricky part is that we have coefficients and
we want to sum the whole thing. I did not try to do reductions (going
from general k tobase case) but since the reductions work in recursive
way the coefficients will be like hypergeometric progression. So
The result seem to be a sum of products of elliptic functions with
some kind of very general hypergeometric functions.

I doubt that one can get more explicit description. OTOH miracles do
happen and due to some strange coincidence the integral may be equal
to something simpler (after all, Axiom got stuck on original integral
in it search for miracles).

-- 
                              Waldek Hebisch
address@hidden